$G$ invariant Riemannian metric on $G/H$ I have been reading a few posts on this site on this particular topic, and I ran into some issues which I want to settle here.
Let $G$ be a Lie group(may assume semisimple if needed) and $H$ its maximal compact subgroup. Let $g =Lie(G), h= Lie(H).$
I've been told that

"Using the Cartan decomposition of $g = h + m$ one can give a $G$-invariant metric on $G/H$"

I want to understand the terms of the above sentence and a proof of it.
1. Here $m$ is supposed to be some sort of orthogonal complement of $h$ in $g$. How does one actually define $m$? Is considering the orthogonal complement wrt Killing form enough? What if we don't start with a semisimple Lie group?
2. To define a Riemannian metric we need a manifold to begin with. Is $G/H$ a manifold?
3. If 3 is true, what is the tangent space of $G/H$ at $eH$? Is it $g/h$? (Seen as the quotient vector space)
4. How does one prove the statement?
Any help is highly appreciated. I hope this will not be called a duplicate question as even though such topics have been discussed, I am asking these questions to clarify some doubts regarding what I've read in the other posts.
 A: The definition of a Cartan decomposition requires that the Lie algebra is semisimple so we don't need to think about other cases.
So firstly we shall take a semisimple Lie group $G$ and its Lie algebra $\mathfrak{g}$.
In the definition of a Cartan decomposition we usually pick an involution $\sigma \in \mathrm{Aut}(\mathfrak{g})$, $\sigma^2=1$ such that $B_\sigma(X,Y) := -B(X,\sigma(Y))$ is a positive definite bilinear form where $B$ is the Killing form.
Then $\mathfrak{h}$ is the $+1$-eigenspace of $\sigma$ and $\mathfrak{m}$ is the $-1$-eigenspace. You can then show that they are orthocomplements with respect to $B$.
$G$ is a manifold and every quotient space $G/H$ by any Lie subgroup inherits a manifold structure naturally. Indeed the tangent spaces are naturally identified with quotients of the Lie algebras: $T_{eH}G/H \cong \mathfrak{g}/\mathfrak{h}$. If you want to see this more explicitly look up the solder form.
The trick in this case is that we have a natural $\operatorname{ad}H$-invariant complement to $\mathfrak{h}$ given by $\mathfrak{m}$. Thus we can identify $\mathfrak{g}/\mathfrak{h} \cong \mathfrak{m}$. Now $B$ (or $B_\sigma$) is nondegenerate and positive definite on $\mathfrak{m}$ giving us a Riemmannian metric on $G/H$.
Note the fact that $B$ is positive definite on $\mathfrak{m}$ follows from the realisation that $\mathfrak{h}$ is maximal compact (compact in Lie algebra terms means $B$ is negative definite on $\mathfrak{h}$).
